Let $c:=\Phi>0$ and $a:=t_{\max}>0$. For any given value (say $m$) of $ET$, the variance $VT$ of $T$ is maximized when $T$ takes only the endpoint values, $0$ and $a$, so that $P(T=a)=m/a=1-P(T=0)$, and then $VT=ma-m^2$ (see the detail on this at the end of this answer). So, 
it remains to note that 
\begin{align}
\max_T ET(ET+cVT)
&=\max_{m\in[0,a]} m(m+c(ma-m^2)) \\
&=\left\{
\begin{alignedat}{2}
&a^2&&\text{ if }ac\le2, \\ 
&\frac{4 \left(a^3 c^3+3 a^2 c^2+3 a c+1\right)}{27 c^2}
&&\text{ if }ac>2.
\end{alignedat}
\right.
\end{align}

**Detail:** If $T$ takes values in $[0,a]$ and $ET=m$, then $ET^2\le a\,ET=ma$, so that $VT=ET^2-(ET)^2\le ma-m^2$. On the other hand, if $P(T=a)=m/a=1-P(T=0)$, then $ET=m$ and $VT=ma-m^2$. So, the variance $VT$ of $T$ is maximized when $T$ takes only the endpoint values, $0$ and $a$.